## Frobenius algebras and linear functionals

Posted: May 5, 2011 in Frobenius Algebras, Noncommutative Ring Theory Notes
Tags: , , ,

Throughout this post $k$ is a field. Recall that if $V$ is a $k$-vector space, then the dual space of $V$ is the vetor spcae $V^*$ consisting of all $k$-linear maps from $V$ to $k.$ If $V$ is finite dimensiona, say with the basis $\{v_1, \ldots , v_n \},$ then for every $1 \leq i \leq n$ we define $e_i \in V^*$ by $e_i(v_i)=1$ and $e_i(v_j)=0$ for all $j \neq i.$ It is easy to see that $\{e_1, \ldots , e_n \}$ is a basis for $V^*.$ In particular, $\dim V = \dim V^*.$

Theorem. Let $A$ be a finite dimensional $k$-algebra. Then $A$ is a Frobenius $k$-algebra if and only if there exists $f \in A^*$ such that $\ker f$ does not contain any non-zero left ideal of $A.$

Proof. Suppose first that $A$ is Frobenius. So there exists a non-degenerate bilinear form $B$ of $A$ such that $B(xy,z)=B(x,yz),$ for all $x,y,z \in A.$ Define $f : A \longrightarrow k$ by $f(x)=B(x,1),$ for all $x \in A.$ Obviously $f \in A^*$ because $B$ is linear. Now suppose that $L$ is a left ideal of $A$ and $L \subseteq \ker f.$ Let $y \in L$ and $x \in A.$ Then $xy \in L$ and so $0=f(xy)=B(xy,1)=B(x,y).$ Therefore $y=0$ because $B$ is non-degenerate. So we have proved that $L=0.$ Conversely, suppose that there exists $f \in A^*$ such that $\ker f$ does not contain any non-zero left ideal of $A.$ Define $B: A \times A \longrightarrow k$ by $B(x,y)=f(xy),$ for all $x,y \in A.$ Clearly $B$ is bilinear because $f$ is linear. Also, clearly $B(xy,z)=f((xy)z)=f(x(yz))=B(x,yz),$ for all $x,y,z \in A.$ To complete the proof of the theorem we only need to show that $B$ is non-degenerate. So suppose that $B(x,y)=0$ for some $y \in A$ and all $x \in A.$ Then $f(xy)=0$ and thus $xy \in \ker f$ for all $x \in A,$ i.e. $Ay \subseteq \ker f.$ Hence $Ay=0,$ because $Ay$ is a left ideal of $A$ and $\ker f$ does not contain any non-zero left ideal of $A.$ So $y=0$ and we are done. $\Box$

Example 1. If $K/k$ is a finite field extension, then $K$ is a Frobenius $k$-algebra.

Proof. Let $\{x_1, \ldots , x_n \}$ be a $k$-basis for $K.$ Define $f : K \longrightarrow k$ by $f(\alpha_1 x_1 + \ldots + \alpha_n x_n)=\alpha_1,$ for all $\alpha_i \in k.$ Clearly $f$ is a non-zero $k$-linear map and $\ker f$ does not contain any non-zero ideal of $K$ because $K$ is a field and so it has no non-zero proper ideal.

Example 2. Let $G$ be a finite group. Then the group algebra $k[G]$ is a Frobenius $k$-algebra.

Proof.  Let $G=\{g_1, \ldots , g_n \},$ where $g_1$ is the identity element of $G.$ Define the map $f : k[G] \longrightarrow k$ by $f(\alpha_1g_1 + \ldots + \alpha_n g_n)=\alpha_1,$ for all $\alpha_i \in k.$ Clearly $f$ is a $k$-linear map and $\ker f =kg_2 + \ldots + kg_n.$ Suppose that $L \subseteq \ker f$ and $L$ is a non-zero left ideal of $k[G].$ Let $0 \neq x= \sum_{i=2}^n \alpha_ig_i \in L.$ Then $\alpha_j \neq 0,$ for some $j \geq 2.$ Since $L$ is a left ideal of $k[G],$ we have $g_j^{-1}L \subseteq L \subseteq \ker f.$ But the coefficient of $g_1$ in $g_j^{-1}x$ is $\alpha_j \neq 0$ and so $g_j^{-1}x \notin \ker f,$ which is a contradiction. Thus $L=0. \ \Box$